Problem: Solve for $x$ and $y$ by deriving an expression for $x$ from the second equation, and substituting it back into the first equation. $\begin{align*}-7x+y &= -3 \\ -3x-y &= -6\end{align*}$
Explanation: Begin by moving the $y$ -term in the second equation to the right side of the equation. $-3x = y-6$ Divide both sides by $-3$ to isolate $x$ $x = {-\dfrac{1}{3}y + 2}$ Substitute this expression for $x$ in the first equation. $-7({-\dfrac{1}{3}y + 2}) + y = -3$ $\dfrac{7}{3}y - 14 + y = -3$ Simplify by combining terms, then solve for $y$ $\dfrac{10}{3}y - 14 = -3$ $\dfrac{10}{3}y = 11$ $y = \dfrac{33}{10}$ Substitute $\dfrac{33}{10}$ for $y$ in the top equation. $-7x+ \dfrac{33}{10} = -3$ $-7x+\dfrac{33}{10} = -3$ $-7x = -\dfrac{63}{10}$ $x = \dfrac{9}{10}$ The solution is $\enspace x = \dfrac{9}{10}, \enspace y = \dfrac{33}{10}$.